Higher-Order Product Formulas

The first-order Trotter formula $e^{-i(A+B)t} \approx e^{-iAt}e^{-iBt}$ is asymmetric, and its error is dominated by a single leftover commutator term at order $t^2$. The symmetric (Strang, or 2nd-order Suzuki) splitting fixes the asymmetry — and cancels that $t^2$ term entirely.

The symmetric splitting

For $H = A + B$ the second-order product formula is

Sandwiching the $B$-evolution between two half-steps of $A$ makes the formula time-reversal symmetric: $S_2(-t) = S_2(t)^{-1}$. That symmetry forces every even power of $t$ in the error expansion (after the leading $1$) to vanish, so the error starts at $O(t^3)$ per step instead of $O(t^2)$:

Suzuki's recursion to higher order

Suzuki showed how to bootstrap $S_2$ into arbitrarily high even orders. The fourth-order formula is built from five fractional $S_2$ steps,

with one step run backwards in time (the middle factor uses a negative coefficient since $1 - 4s < 0$). The pattern continues: $S_{2k}$ is assembled recursively from $S_{2k-2}$, giving error $O(t^{2k+1})$ per step at the cost of an exponentially growing (in $k$) number of exponentials.

Mapping exponentials to rotation gates

On a single qubit, the Pauli exponentials are rotation gates. With the convention $R_P(\theta) = e^{-i\theta P/2}$ used by this simulator,

Take the toy Hamiltonian $H = X + Z$ (so $A = X$, $B = Z$). The symmetric step $S_2(t)$ becomes

Each half-step $e^{-iX t/2}$ is $R_X(t)$ (because $R_X(t) = e^{-iX t/2}$), and the full $B$-step $e^{-iZ t}$ is $R_Z(2t)$.

Try it

Build one symmetric 2nd-order Trotter step for $H = X + Z$ at $t = 0.4$. Apply $R_X(t)$, then $R_Z(2t)$, then $R_X(t)$ on the single qubit. The grader checks the full unitary, so the angles and the symmetric ordering must be exact.

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